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\documentclass{article}\usepackage{graphicx} % Required for inserting images\usepackage{hyperref}\usepackage{asmmath} % Problem happens here\title{The equations that I know}\author{Made by Erik Andrade Lins}\date{July 2023}\begin{document}\maketitle\begin{center}Version 1\end{center}\section{Introduction}In this file made with this \href{https://www.overleaf.com}{\LaTeX\,editor}, I'll be showing you some equations that I know.\section{The equations}\subsection{Fourier transforms}Credits to \href{https://www.youtube.com/watch?v=spUNpyF58BY}{3Blue1Brown} for the fourier transform formulas.\subsubsection{Pseudo-fourier}\[\hat{g}(f) = \frac{1}{{t_2 - t_1}}\int_{t_1}^{t_2} g(t)e^{-2\pi\,ift}\,dt\]\subsubsection{Actual}This one is exactly the same as the one on section 2.1.1 except that the fraction $\frac{1}{t_2 - t_1}$ is not multiplied to the integral. The equation is:\\\[\hat{g}(f) = \int_{t_1}^{t_2} g(t)e^{-2\pi\,ift}\,dt\]\subsection{Exponentiating $e$}\subsubsection{Taylor's expansion of $e^{x}$}I credit \href{https://youtu.be/B1J6Ou4q8vE?t=454}{Alan} for this one. The equation is this:\\\[e^{x} = \sum\limits_{n=0}^{\infty} \frac{x^{n}}{n!}\]\subsubsection{Euler's formula for $e^{ix}$}Also crediting \href{https://youtu.be/B1J6Ou4q8vE?t=256}{Alan}. The equation is $e^{ix} = \cos\,x + i\sin\,x$. Did you know that $e^{i\pi} = -1$?\subsection{Newton's law of force}$F$ is the force you require to move something of mass $m$ on a friction-less surface where the acceleration you want is $a$. The equation is $F = ma$, though you could write it as $F = am$.\subsection{Exchange for mass and energy}Everyone knows this equation. It was developed by Einstein and it's $E = mc^{2}$.\subsection{Disproofs of commutation on division, exponentiation and subtraction}Please note that the disproofs are no longer true if the equation $x = y$ is satisfied, though for division you need to be careful because if you define $x$ and/or $y$ to be $0$, you'll be handling with a division by $0$, which is undefined (It's exactly why $\int_{0}^{\infty} t^{x}e^{-t} \,dt$ has asymptotes. You have to divide by $0$ every time $x$ is a negative number to satisfy $\int_{0}^{\infty} t^{x}e^{-t} \,dt = x\int_{0}^{\infty} t^{x-1}e^{-t} \,dt$. So the inverse is $\int_{0}^{\infty} t^{x-1}e^{-t} \,dt =\frac{\int_{0}^{\infty} t^{x}e^{-t} \,dt}{x}$. $x$ in this case is $0$ so, it means to get $\int_{0}^{\infty} t^{-1}e^{-t} \,dt$, we have to divide by zero. If you wanted to know the exact fraction, here it is: $\frac{1}{0} = \int_{0}^{\infty} t^{-1}e^{-t} \,dt$ ($\int_{0}^{\infty} t^{0}e^{-t} \,dt$ is $1$).).\subsubsection{Division}$\frac{x}{y} \neq \frac{y}{x}$. We can prove this using proof by contradiction. Let's suppose that $\frac{x}{y} = \frac{y}{x}$, $x = 1$ and that $y = 2$. $\frac{1}{2}$ is $0.5$, but $\frac{2}{1}$ is $2$, implying that $\frac{1}{2} \neq \frac{2}{1}$ and that $\frac{1}{2} - \frac{2}{1} \neq 0$, disproving $\frac{x}{y} = \frac{y}{x}$.\subsubsection{Exponentiation}$x^{y} \neq y^{x}$. Again, we can prove this using proof by contradiction. Assume again that $x^{y} = y^{x}$, $x = 1$ and $y = 2$. $1^{2} = 1$, but again, $2^{1} = 2$, suggesting again that $1^{2} \neq 2^{1}$ and that $1^{2} - 2^{1} \neq 0$, again disproving $x^{y} = y^{x}$.
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